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3 years ago

The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i$. What is $a+b$?

1 answer:

6 0

We know for our problem that the zeroes of our quadratic equation are $(2-3i)$ and $(2+3i)$ , which means that the solutions for our equation are $x=2-3i$ and $x=2+3i$ . We are going to use those solutions to express our quadratic equation in the form $a x^{2} +bx+c$ ; to do that we will use the zero factor property in reverse:
 $x=2-3i$
$x-2=-3i$
$x-2+3i=0$

 $x=2+3i$
$x-2=3i$
$x-2-3i=0$

Now, we can multiply the left sides of our equations:
 $(x-2+3i)(x-2-3i)= x^{2} -2x-3ix-2x+4+6i+3ix-6i-3^2i^2$
= $x^{2}-4x+4-9i^{2}$
= $x^{2} -4x+4+9$
= $x^{2} -4x+13$
Now that we have our quadratic in the form $a x^{2} +bx+c$ , we can infer that $a=1$ and $b=-4$ ; therefore, we can conclude that $a+b=1+(-4)=1-4=-3$ .

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Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

3 years ago

Write an equation in the form y=mx for the proportional relationship that passes through the points (2, -15) and (6, -45).

nydimaria [60]

Answer:

Y=-15/2x

Step-by-step explanation:

7 0

3 years ago

Find the complex zeros of x^3+27. write f in factored form

aliina [53]

$x^3+27=0
\\x^3+3^3=0
\\(x+3)(x^2-3x+9)=0
\\x^2-3x+9=0
\\a=1,b=-3,c=9
\\
\\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} 
\\
\\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0

$

4 0

3 years ago

he temperature at noon on a winter day was 2 °F. By 6 p.m., the temperature had dropped to −9 °F. Write a subtraction expression

nydimaria [60]

Based on the starting temperature and the new temperature, the relevant expression would be 2 - (-9)

To find the number of degrees the temperature has changed by, use the following equation:

= Starting temperature - Current temperature

Starting temperature = 2°F

Current temperature = -9°F

The equation becomes:

= 2 - (-9)

In conclusion, the expression would be 2 - (-9)

Find out more on expressions at brainly.com/question/945593.

5 0

2 years ago

Tina thinks of two numbers. She says, "My two numbers add up to 12. Three times my first number plus twice my second number is 2

butalik [34]

The first number is 5 and the second number is 7

8 0

3 years ago